Problem: $f'(x)=-5f(x)$, and $f(0)=4$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $f(x)=4+e^{-5x}$ (Choice B) B $f(x)=e^{-5x}$ (Choice C) C $f(x)=4e^{-5x}$ (Choice D) D $f(x)=3+e^{-5x}$
The general solution of equations of the form $f'(x)=kf(x)$ is $f(x)=C\cdot e^{kx}$ for some constant $C$. This can be found using separation of variables. In our case, $k=-5$, so $f(x)=C\cdot e^{-5x}$. Let's use the fact that $f(0)=4$ to find $C$ : $\begin{aligned} f(x)&=C\cdot e^{-5x} \\\\ f(0)&=C\cdot e^{-5\cdot 0} \gray{\text{Plug }x=0} \\\\ 4&=C\cdot e^{-5\cdot 0} \gray{f(0)=4} \\\\ 4&=C \end{aligned}$ In conclusion, $f(x)=4e^{-5x}$.